AP Calculus AB and BC

Limit of ln x as x Approaches Infinity

The limit of ln x as x approaches infinity is infinity. The logarithm has no ceiling, so its graph has no horizontal asymptote on the right. What makes it feel bounded is the pace: to push ln x past 100 you need an input larger than 10 to the 43rd power.

\lim_{x \to \infty} \ln x = \infty

Settled by end behaviour of the logarithm.

Reading the growth off the exponential

$\ln x$ is the inverse of $e^{x}$ , so $\ln x = y$ and $e^{y} = x$ are the same statement. Asking how large $\ln x$ can get is asking which exponents $e^{y}$ can reach, and $e^{y}$ reaches every positive number.

Pick any bound $M$ . The input $x = e^{M}$ has $\ln x = M$ , and every larger input gives a larger logarithm because $\ln$ is increasing. No bound survives, so the values are unbounded above.

\lim_{x \to \infty} \ln x = \infty

$x$	$\ln x$
$10$	$2.303$
$10^{3}$	$6.908$
$10^{6}$	$13.816$
$10^{43}$	$99.011$

Each step down that table multiplies the input by a thousand and adds only about $6.9$ to the output. The climb never stops and it never speeds up, and that combination is the source of nearly all confusion about this limit.

Why substitution fails, and why nothing is indeterminate

There is no number to substitute. $\infty$ describes behaviour rather than naming a value, so $\ln \infty$ is shorthand for the question and not an answer to it.

No indeterminate form appears afterwards either. Shapes like $\frac{\infty}{\infty}$ and $\infty - \infty$ arise when two quantities pull against each other. One increasing function walking out to the right end of its domain has no opponent, so end behaviour alone settles the value.

Slow is not the same as bounded

Growth that keeps decelerating can still be unbounded. $\ln x$ , $\sqrt{x}$ , and $\sqrt[3]{x}$ all flatten out visually and all run to $\infty$ . Contrast $\arctan x$ , which genuinely stops at $\frac{\pi}{2}$ because its range forbids anything higher.

The slowness does matter once something else is in the expression. Against any positive power of $x$ the logarithm loses, which is why $\lim_{x \to \infty} \frac{\ln x}{x} = 0$ even though both parts are heading to $\infty$ .

The mistake students make

Claiming a horizontal asymptote because the graph looks flat on a calculator screen. A window shows finitely much, and the values keep rising past every level in it.
Confusing the two ends. $\ln x \to \infty$ as $x \to \infty$ , while $\ln x \to -\infty$ as $x \to 0^{+}$ . Both ends are unbounded, in opposite directions.
Assuming the logarithm catches a power somewhere. $\frac{\ln x}{x^{p}} \to 0$ for every $p > 0$ , so any positive power takes over eventually.
Applying L'Hopital's rule to $\ln x$ by itself. The rule needs a quotient in an indeterminate form, and a lone unbounded function is neither.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Does ln x have a horizontal asymptote?

No. A horizontal asymptote needs a finite limit at an infinite end, and neither end of $\ln x$ is finite. It does have a vertical asymptote, the line $x = 0$ , where the values fall to $-\infty$ .

Does the base change the answer?

Not for any base above 1. Changing base multiplies by a positive constant, since $\log_b x = \frac{\ln x}{\ln b}$ , and a positive constant times an unbounded increasing function is still unbounded. For a base between 0 and 1 that constant is negative and the limit becomes $-\infty$ .

Why do so many limits containing ln x still come out finite?

Because the logarithm is usually paired with something stronger. $\frac{\ln x}{x} \to 0$ at infinity and $x \ln x \to 0$ as $x \to 0^{+}$ . A power of $x$ outruns a logarithm at both ends, so the logarithm decides the answer only when nothing is competing with it.